Physics I- Classical Mechanics with an Experimental Focus- Jumping from a Height

3 Pages·0.05 MB·English

by  Peter Dourmashkin;Kate Scholberg

MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
Physics 8.01X Fall Term 2002
JUMPING FROM A HEIGHT
(cid:1)
(cid:0)(cid:1)
(cid:0)
h
(cid:6)
(cid:6) D
(cid:6) h
(cid:6)
(cid:3) (cid:6)
(cid:2)(cid:3) (cid:6)
(cid:2) (cid:5)
(cid:4)(cid:5)
(cid:7) (cid:4)
(cid:7)
(cid:7)
(cid:7)
(cid:7)
t=0 t=t1 t=t2
In this example we will consider what happens if you bend your knees
when you hit the ground if you are jumping from a height.
Imagine you have mass m and jump from height h at time t = 0. You
hit the ground at time t = t . Then between t = t and t = t over interval
1 1 2
¢t = t t you bend your knees and lower your center of mass by a distance
2 1
¡
¢h. In the flgure the red spot indicates your center of mass.
We will ask the questions:
a. What is the average force of the ground on your legs during the
impact, in terms of m, g, h and ¢h?
b. What is ¢t over which the impact happens, in terms of g, h and
¢h?
c. If your mass is m = 60 kg, what is the maximum h=¢h that you
can sustain without breaking your tibia? We will assume that the
compressive force per area necessary to break the tibia in the lower
leg is about 1.6 103 bars (1 bar = 105 Pa = 105N/m2). The smallest
£
cross-sectional area of the tibia, about 3.2 cm2, is slightly above the
ankle.
For this example, we’ll consider you to be a point particle located at your
center of mass (we’ll see a bit later in the class why this is reasonable).
To solve part (a), flrst let’s consider the flrst time interval from t = 0 to
t = t , during which you are falling. You can get your flnal velocity at the
1
end of this interval from either 1D kinematics or the work-energy theorem.
Let’s do the latter. We’ll assume no friction, so there’s no non-conservative
work, so
¢KE +¢PE = 0
1
mv2 mgh = 0 (1)
2 1 ¡
so
v = 2gh
1
q
.
Now let’s consider the time interval from t = t to t = t , during the
1 2
impact. We can apply the work-energy theorem again,
¢KE +¢PE = W
NC
But this time, there is non-conservative work being done on you! There
is a contact force from the (cid:176)oor on you, and its direction is antiparallel to
your displacement. So the contact force does negative work on you, and
this non-conservative work is W = F ¢h. (We’ll assume this force is
NC floor
¡
constant over the interval).
The work-energy theorem for this interval gives:
1
mv2 mg¢h = F ¢h
¡2 1 ¡ ¡ floor
(where the \¢" refers to before and after the impact. Before, you have
velocity v . After, you have come to a stop and you have velocity zero. Your
1
change in P.E. over the interval is mg¢h.)
¡
Now we can plug in 1mv2 = mgh from equation1, and rearrange terms a
2 1
little to get
F = mgh+mg¢h
floor ¢h
So that’s part (a).
Now for part (b), let’s flnd the impulse in order to flnd ¢t, since impulse
is average total force times time. Impulse is also change in momentum, ¢p.
We know that your change in momentum over the interval is ¢p = p p =
2 1
¡
0 mv = mv .
1 1
¡ ¡
¢p = F ¢t
tot
(which is another way of stating Newton’s 2nd Law), where F is the
tot
total average force on you. There are two forces acting on you: gravity and
the force of the (cid:176)oor. So F = F mg, where F is the average force
tot floor floor
¡
of the (cid:176)oor that we just calculated in part (a).
So, the magnitude of ¢t is then j¢pj
jFtotj
Plugging in:
¢p mv
1
¢t = j j =
F mg mgh+mg¢h mg
floor
j ¡ j ¢h ¡
The m’s cancel, and plugging in for v and massaging a little more we
1
get:
p2gh
¢t =
g(h=¢h)
For part (c), we use our expression for F to get
floor
h
F = mg +1
floor
ˆ¢h !
Themaximumforcethatthesmallestareaofthetibiacantakeis1:6 103
£
bars times 105 N/m2 per bar times 3:2 10¡4 m2 times 2 (for two legs) is
£
F = 1:0 105 N. If we take F = F when the tibia just breaks, and
max max floor
£
solving for h=¢h, we get (h=¢h) = 173 .
max
So, if you don’t bend your knees (take ¢h = 1 cm), you will break your
legs jumping from only 1.7 m. If you bend your knees 0.5 m, your leg bones
may survive a leap from 87 m! (Please don’t try this yourself though!! This
problem considers only damage to bones{ in fact other tissues in your body
could get damaged in a fall from a height of more than a few meters).
And if you are falling into something soft and cushiony, or into water, ¢h
(and ¢t) are relatively larger. Parachutists are trained to maximize time and
displacement of impact when landing by crouching and rolling. And compare
a dive to a belly-(cid:176)op: small ¢h and ¢t during the collision hurts more!
)